Answer: 41.5 mL
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,
n = moles of solute
= volume of solution in L
Given : 59.4 g of
in 100 g of solution
moles of 
Volume of solution =
Now put all the given values in the formula of molality, we get

To calculate the volume of acid, we use the equation given by neutralisation reaction:

where,
are the molarity and volume of stock acid which is 
are the molarity and volume of dilute acid which is 
We are given:

Putting values in above equation, we get:

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid
Answer:
Explanation:
The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below
H
|
H₂N - C - C =O
| \
H OH
The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.
Answer:
41 g
Explanation:
We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.
pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]
pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]
log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]
log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40
[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M
We can find the mass of NaC₆H₅COO using the following expression.
M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L
mass NaC₆H₅COO = 41 g
Hello!
Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).
We have the following data:
m (mass) = ?
n (number of moles) = 5.20 moles
MM (Molar mass of C6H12) ≈ 84.2 g/mol
Now, let's find the mass, knowing that:




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I Hope this helps, greetings ... Dexteright02! =)